So the Mean Value Theorem states that if you are given a function that is continuous and differentiable (which just means no jumps, holes, or corners) the slope of the secant line and the tangent line are the same at point "c". This means that they are two parallel lines.
Alright, so let's try this out. We will be using a very familiar graph to make things a litttle easier to see and understand. (At least i worked for me)
We have the graph f(x)= x² on [0,2]
So a=0 and b=2. Let's get this numbers in the formula [f (b)-f (a)]/(b-a)
1) [f(2)-f(0)]/(2-0)= 4-0/2=2 so m=2
2) Now we will replace x by the two we just got into the f(x)>>>>f(2)=4
Now we have a point and a slope so let's use the point-slope formula
3) y-4= 2(x-2) which is y=2x. This is our secant equation. (The green line on the graph)
We know the f '(x) = 2x so f '(c) =2cWe also know that 2c= slope and we know the slope is 2. Therefore c=1.
Now we find out what f(c) equals when c=1>>>>>> its 1 so we now know the tangent line has to go throught the point (1,1) with the slope 2x. We do the point-slope formula.
4) y-1= 2(x-1)
y-1=2x-2
y= 2x-1 (It's the red line)
So why doesnt this work on a function that's not smooth? Let me show you with an example AKA another graph.
As you can see the tangent line doesn't exist at x=0
